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Question:

There is a tunnel connecting city A and B. There is a CAT which is standing at 3/8 the length of the tunnel from A. It listens a whistle of the train and starts running towards the entrance where, the train and the CAT meet. In another case, the CAT started running towards the exit and the train again met the CAT at the exit. What is the ratio of their speeds?

Explanation:

Solution:

Let the speed of the train be St and the speed of the CAT be S_c.

Case 1:

The CAT runs towards the train.

$\frac{d}{S_{t}} = \frac{3 x}{8 S_{c}}$

$∴ \frac{3 x}{8} \times \frac{S_{t}}{S_{c}} = d$ ...(i)

Case 2:

The CAT runs away from the train.

$\frac{d + x}{S_{t}} = \frac{5 x}{8 S_{c}}$

$∴ d = \frac{5 x}{8} \times \frac{S_{t}}{S_{C}} - x$ ...(ii)

Equating (i) and (ii),

$\frac{3 x}{8} \times \frac{S_{t}}{S_{c}} = \frac{5 x}{8} \times \frac{S_{t}}{S_{c}} - x ∴ \frac{3}{8} x \times \frac{S_{t}}{S_{c}} = \frac{5 x}{8} \times \frac{S_{t}}{S_{c}} - x ∴ \frac{3}{8} \frac{S_{t}}{S_{c}} = \frac{5}{8} \frac{S_{t}}{S_{c}} - 1$

∴ S_t : S_c = 4 : 1

Hence, option (a).

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